If X is a Poisson variate such that 4P(X=1)=3P(X=2), then P(X=0)= Option 1: e^(-8/3) Option 2: e^(-4/3) Option 3: e^(-2) Option 4: e^(-1/3)

Answer: e^(-8/3). Solution: With mean, 4 e^(-)=3*(^2)/2e^(-), so 8=3 and =83. Hence P(X=0)=e^(-)=e^(-8/3).

Applied Mathematics

Probability Distribution

Medium

applied

If $X$ is a Poisson variate such that $4P(X=1)=3P(X=2)$ , then $P(X=0)=$

$e^{-8/3}$

$e^{-4/3}$

$e^{-2}$

$e^{-1/3}$

✅ Correct Option: 1

With mean $\lambda$ , $4\lambda e^{-\lambda}=3\cdot\dfrac{\lambda^2}{2}e^{-\lambda}$ , so $8=3\lambda$ and $\lambda=\dfrac83$ . Hence $P(X=0)=e^{-\lambda}=e^{-8/3}$ .

2023: 22 May Shift 3

If the probability distribution of X is :

X	2	3	4	5	6
P(X)	1/15	2/15	3/15	4/15	5/15

Then variance is equal to :

2026: 26 May Shift 2

A traffic police officer records the number of motorcycle riders that use a particular lane on a highway. He records that an average of 2.5 motorcycle riders use the lane every hour. Given that the number of motorcycles that use the particular lane follows a Poisson distribution. The probability that less than 2 motorcycle riders will use the lane within an hour will be:- (Use $e^{2.5} = 12.2$ )

2023: 23 May Shift 3

A book consisting of 2000 pages has 540 misprints distributed randomly throughout the book. The average number of misprints in one page of the book is:

2026: 30 May Shift 2

Match List-I with List-II

List-I	List-II
(A) Mean of a normal variate X is 12 and standard deviation is 4, then the Z-score of data point 20 is	(I) 1
(B) If X is Poisson variate such that P(X=1) = 2P(X=2), then the mean of X is	(II) 8
(C) In a binomial distribution, if mean is 5 and variance is 4, then the number of trials is	(III) 2
(D) In a binomial distribution, if n = 20 and q = 0.6, then its mean is	(IV) 25

Choose the correct answer from the options given below:

2026: 29 May Shift 1

A random variable X has the following probability distribution:

X	0	1	2	3	4
P(X)	0.1	0.15	0.30	0.30	0.15

Which of the following are correct?

(A) $P(X \le 1)=0.25$

(B) $P(X>3)=0.15$

(D) $P(X \le 2)=0.55$

Choose the correct answer from the options given below:

2023: 23 May Shift 3

If the mean of a binomial distribution is 12 and its standard deviation is 2, then the number of trials is :

If XXX is a Poisson variate such that 4P(X=1)=3P(X=2)4P(X=1)=3P(X=2)4P(X=1)=3P(X=2), then P(X=0)=P(X=0)=P(X=0)=

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $X$ is a Poisson variate such that $4P(X=1)=3P(X=2)$ , then $P(X=0)=$